Rocket staging — series staging, parallel staging: Visual walkthrough

3.3.45 · D2 · Physics › Rocket Propulsion › Rocket staging — series staging, parallel staging

Yeh page parent note Rocket staging ka visual companion hai. Yeh Tsiolkovsky rocket equation aur Structural fraction and propellant fraction par rely karta hai.

Step 1 — Ek rocket actually kya hota hai: mass ke teen pile

KYUN. Har rocket formula ek kahani hai — kaunsa pile rehta hai aur kaunsa pile chala jaata hai. Agar hum abhi piles ko alag nahi karte, toh hum "exhaust ke roop mein nikalne wala fuel" ko "baad mein drop hone wala empty tank" se confuse kar lenge — aur ye dono cheezein hamaari speed par bilkul alag kaam karti hain. Isliye hum pehle inhe alag draw karte hain.

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Figure — Rocket staging — series staging, parallel staging

Step 2 — Rockets ka ek rule: mass peeche fenko, aage push pao

KYUN $v_{e}$ use karen aur "thrust" ya "force" nahi? Force har second change hoti hai jaise rocket halka hota hai, jisse use add up karna awkward ho jaata hai. Lekin exhaust speed $v_{e}$ ek given engine ke liye roughly constant rehti hai (yeh Specific impulse se set hoti hai). Ek constant number ke saath reason karna kaafi aasaan hota hai — isliye hum sab kuch $v_{e}$ par build karte hain. Typical chemical engines ka $v_{e}$ kuch hazaar metres per second hota hai; hum baad mein realistic values jaise $3000$ – $3500$ m/s use karenge.

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Step 3 — Logarithm kyun aata hai (ek million tiny kicks ko add karna)

KYUN logarithm aur simple multiplication nahi? Kyunki har kick pichle wale se choti hoti hai — start mein rocket heavy hota hai (zyada mass move karna padta hai) aur end mein halka. Step 2 ko rearrange karne par $d v = - v_{e} \frac{d m}{m}$ milta hai. Quantity $\frac{d m}{m}$ mass ka ek fractional change hai, aur $m_{0}$ se $m_{f}$ tak ke fractional changes ko sum karne par $ln (m_{0} / m_{f})$ milta hai. Natural log $ln$ precisely "fractional changes ka running total" hai.

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Step 4 — Trap: ek empty tank carry karna dead weight hai

KYUN yeh matter karta hai. Mass ratio log ke andar hai, aur log jaldi flatten hota hai — ratio ko double karne se speed double nahi hoti (Step 3 mein ghatta hua area phir dekho). Toh heavy dead $m_{f}$ ki sakht saza milti hai. Yeh parent note ka "vicious cycle" hai, ek picture ke roop mein dekha gaya.

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Step 5 — Staging ka idea: tank drop karo, ratio reset karo

KYUN yeh jeet jaata hai. Har stage ab ek fresh, favourable mass ratio dekhta hai, kyunki yeh kabhi bhi ek spent tank carry nahi karta. Do logs multiply nahi hote — wo add hote hain, kyunki $Δ v$ ek speed hai aur speeds accumulate hoti hain. Do decent ratios mile hue ek crippled ratio ko beat kar dete hain.

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Burnout ke baad hum $m_{s, 1}$ ko jettison karte hain — woh gray block gir jaata hai aur stage 2 ke numbers mein kabhi nahi aata.

Step 6 — Stage 2 ek cleaner ratio dekhta hai; speeds add hoti hain

KYUN add hota hai multiply nahi? Stage 2 already $Δ v_{1}$ par move kar raha hota hai (velocity relative hoti hai; rocket separation par apni speed "bhool" nahi jaata). Naya burn $Δ v_{2}$ ko upar se add karta hai. Total speed sum hai.

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Step 7 — Numbers daalo (worked two-stage rocket)

KYUN ye particular numbers? Exhaust velocities realistic chemical-engine values hain (Step 2 ne few-thousand-m/s range flag kiya tha): upper stage $v_{e, 2} = 3500$ m/s use karta hai (upper stages vacuum mein run karte hain, jahan nozzles zyada efficient hote hain → zyada $v_{e}$ ), jabki lower stage $v_{e, 1} = 3000$ m/s use karta hai (yeh dense atmosphere se ladhta hai, isliye thoda kam). Propellant fractions Structural fraction and propellant fraction se aate hain.

KYUN top-down? Ek lower stage ka "payload" uske upar ki sab cheez hoti hai. Tum tab tak nahi jaante ki stage 1 ko kya lift karna hai jab tak tumhe stage 2 ka total hardware mass $S_{2}$ nahi pata. Isliye hum tip se start karte hain aur neeche aate hain.

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Worked example Two-stage series rocket

Diya gaya. Final payload $m_{pl} = 1000$ kg; $v_{e, 1} = 3000$ m/s, $v_{e, 2} = 3500$ m/s; $λ_{2} = 0.87, ϵ_{2} = 0.13$ ; $λ_{1} = 0.85, ϵ_{1} = 0.15$ .

Stage 2 — hardware choose karo $S_{2} = 8000$ kg:

$m_{p, 2} = λ_{2} S_{2} = 0.87 \times 8000 = 6960$ kg, $m_{s, 2} = ϵ_{2} S_{2} = 0.13 \times 8000 = 1040$ kg
$m_{0} = S_{2} + m_{pl} = 8000 + 1000 = 9000$ kg, $m_{f} = m_{s, 2} + m_{pl} = 1040 + 1000 = 2040$ kg $Δ v_{2} = 3500 ln (\frac{9000}{2040}) = 3500 \times 1.4839 \approx 5194 m/s$

Stage 1 — hardware choose karo $S_{1} = 90000$ kg; ise stage 2 + payload $= S_{2} + m_{pl} = 8000 + 1000 = 9000$ kg lift karna hai:

$m_{p, 1} = λ_{1} S_{1} = 0.85 \times 90000 = 76500$ kg, $m_{s, 1} = ϵ_{1} S_{1} = 0.15 \times 90000 = 13500$ kg
$m_{0} = S_{1} + 9000 = 99000$ kg, $m_{f} = m_{s, 1} + 9000 = 22500$ kg $Δ v_{1} = 3000 ln (\frac{99000}{22500}) = 3000 \times 1.4816 \approx 4445 m/s$

Total: $Δ v \approx 5194 + 4445 = 9639 m/s$ — kaafi hai (with Gravity losses budgeted) low Earth orbit ke liye.

Step 8 — Edge cases: agar hum stage na karein, ya bura stage karein toh?

Case A — ek stage, kabhi tank drop nahi kiya. Toh $m_{f}$ mein hamesha poora structure hoga. Jaise tum propellant add karte ho, $m_{0} / m_{f}$ badhta hai lekin saturate hota hai: log flatten hota hai, $Δ v$ ek ceiling hit karta hai. Yeh Step-4 trap numbers mein hai.

Case B — zero structure ( $ϵ \to 0$ ). Ek magical weightless tank. Tab $m_{f} \to m_{pl}$ aur ratio bahut bada ho sakta hai — woh ideal jo har engineer Payload fraction optimization ke zariye chase karta hai. Unreachable hai, lekin yeh dikhata hai ki structure dushman hai.

Case C — saari structure, koi fuel nahi ( $λ \to 0$ ). Tab $m_{0} = m_{f}$ , isliye $m_{0} / m_{f} = 1$ aur $ln 1 = 0$ : zero speed. Koi propellant na ho wala stage kuch nahi karta — yeh pure dead weight hai. Yeh degenerate floor hai.

Case D — infinitely many tiny stages. Structure ko continuously drop karna theoretical maximum dega, lekin har separation ek failure point hai, isliye real rockets 2–3 stages use karte hain (Falcon Heavy, Saturn V).

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Ek-picture summary

Recall Feynman retelling — poora walkthrough simple shabdon mein

Ek rocket teen pile hai: fuel, empty structure, aur payload. Yeh fuel ko fixed speed $v_{e}$ par peeche fenk kar fast hota hai. Fuel ka har chhota scoop ek chhota kick deta hai; jab tum saare kicks add karte ho (woh integral $m_{0}$ se $m_{f}$ tak) toh tumhe mass ratio ka logarithm milta hai — Tsiolkovsky, $Δ v = v_{e} ln (m_{0} / m_{f})$ . Problem: log jaldi flatten hota hai, aur $m_{f}$ mein baitha ek empty tank tumhara ratio choke karta hai — $10%$ structure ke saath tum ratio of $10$ se zyada nahi ja sakte, jo tumhe orbital speed se neeche cap kar deta hai. Toh hum cheat karte hain: rocket ko stacked stages mein banao, ek jalao, phir uska dead tank drop karo taaki next stage halka start kare. Har clean stage apna healthy log get karta hai, aur kyunki velocities sirf add hoti hain, do logs ek bade total mein stack ho jaate hain. Top-down kaam karte hue (stage 2 sirf payload $m_{pl}$ carry karta hai; stage 1 stage 2 ka hardware $S_{2}$ bhi carry karta hai), $v_{e, 1} = 3000$ aur $v_{e, 2} = 3500$ wale hamare example mein lagbhag 5.2 km/s + 4.4 km/s ≈ 9.6 km/s milta hai — orbit. Aur edge cases humein honest rakhte hain: koi fuel nahi matlab $ln 1 = 0$ (kuch nahi), koi structure nahi matlab bahut bada ratio (sapna), aur kabhi tank na drop karna matlab tum ek aisi wall se takraate ho jo koi bhi fuel nahi tod sakta.

Recall Quick self-test

Stage $Δ v$ 's multiply kyun nahi hote, add kyun hote hain? ::: Kyunki $Δ v$ ek velocity hai; rocket separation ke through apni speed keep karta hai, isliye har burn apna gain pichle wale ke upar add karta hai. Staging single stage ko kyun beat karta hai? ::: Empty structure drop karna har stage ke liye $m_{f}$ small rakhta hai, isliye har mass ratio (aur uska log) bada rehta hai. Koi propellant na ho wale stage ka $Δ v$ kya hoga? ::: Zero — $m_{0} = m_{f}$ se $ln (1) = 0$ milta hai. $S_{i}$ ka matlab kya hai? ::: Stage $i$ ka hardware mass, $S_{i} = m_{p, i} + m_{s, i}$ (uska apna propellant plus structure, jo payload woh carry karta hai usse count nahi karte).